What is the arc length if Θ = 2 pi over 7 and the radius is 3 cm?
7 pi over 12 12 pi over 7 6 pi over 7 7 pi over 6
@campbell_st
well isn't arc length \[l = \theta \times r\] so multiply that angle by the radius
How is it not arc length? So, it would be L= 2pi/7 x 3?
well I said..."isn't arc length"
that's correct... you just need to simplify it
yeah, im not really good at that :( sorry, it never caught on
thats ok.. so what answer do you think...?
I guessing 6pi/7 ?
great guess
lol, yaay! thank you
glad to help
Question though, how do you solve that? Like changing a radian into a degree and vice versa?
assuming the angle is in radians.... you should say so. r =3cm, diameter = 6 cm and circum = 6π arc length = circum(angle/2π) .......use this formula to get your answer...........
ok... so you need to know \[180^o = \pi\] so \[1^o = \frac{\pi}{180}\] and with radians \[1^r = \frac{180}{\pi}\] so as a simple explanation... an equalateral triangle has all sides equation and all angles 60 imagine a sector when the 2 radii and the arc are the same length... its like pushing on 2 vertices of an equlilateral and bending the 3rd side to make an arc... the angle goes from 60 to a value a bit less than 60... and its called a radian... so 1 radian = 57.29 degrees.... so by bending the side out... the angle has to get smaller.. hope that's not 2 confusing
kind of, yeah.
well just remember 180 = pi.... and then just manipulate that
Ok, so if I have 90 degrees and want to change it into a radian I would use 180/pi?
so divide both sides by 2 \[\frac{180}{2} = \frac{\pi}{2} ...or.... 90^o = \frac{\pi}{2}\]
45?
ok.... so 45 is half of 90.. so divide the previous by 2 \[\frac{90}{2} = \frac{\frac{\pi}{2}}{2}.... or...... 45 = \frac{\pi}{4}\]
11.25?
wait, is pi/4 the answer?
thats it... where ever possible leave the angle in terms of pi
oh ok! Thanks for explaining, I'm a little rusty but I think I get it
Can you help with another question? It's about amplitude and periods etc
ok... just post it... as a new question... and there are plenty of people who can help
ok thanks.
Thanku soo much!! :D
@malice ;)
@malice this is how i found u hehe
LOL cool haha :D
I can't believe that I have already finished this course... *sniff* ^-^
Life really does go on.............
@malice im still finishing it
Join our real-time social learning platform and learn together with your friends!